User aleksandar bahat - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T09:30:35Z http://mathoverflow.net/feeds/user/25791 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/124719/where-was-the-arithmetic-zeta-function-of-a-scheme-first-defined Where was the arithmetic zeta function of a scheme first defined? Aleksandar Bahat 2013-03-16T19:13:14Z 2013-03-16T19:19:44Z <p>Let $X$ be an arithmetic scheme, that is, a scheme of finite type over the integers. We denote the set of closed points of $X$ by $|X|$. For every $x\in|X|$, write $N(x)$ for the cardinality of the residue field $\kappa(x)$.</p> <p>The <em>arithmetic zeta function</em> of $X$ is defined as $$\zeta_X(s)=\prod_{x\in|X|}\frac{1}{1-N(x)^{-s}}.$$</p> <p>This definition (up to a change in variable) can be found in</p> <blockquote> <p>A. Grothendieck, Formule de Lefschetz et rationalité des fonctions $L$, <em>Séminaire Bourbaki</em> 279 (1964), 41-55.</p> </blockquote> <p>Grothendieck attributes this definition to Weil, but as far as I know, Weil only defined the <em>Hasse-Weil zeta function</em>: if $X$ is a smooth projective variety over $\mathbb{F}_q$ and $N_r=|X(\mathbb{F}_{q^r})|$, then $$Z_X(t)=\exp\left(\sum_{r=1}^\infty N_r(X)\frac{t^r}{r}\right).$$</p> <p>Of course, it is easy to show these two functions satisfy $$\zeta_X(s)=Z_X(q^{-s}),$$ but Weil did not address the notion of the zeta function of a scheme, at least not in the original paper:</p> <blockquote> <p>A. Weil, Numbers of solutions of equations in finite fields, <em>Bull. Amer. Math. Soc.</em> 55 (1949), 497-508.</p> </blockquote> <p>Where was the zeta function <em>of an arithmetic scheme</em> first defined? If anyone knows the actual paper in which this first appears, that would be optimal. (Of course, it's very possible that this definition was well-known but unpublished for some time; that would be an acceptable answer too.)</p> http://mathoverflow.net/questions/120067/what-do-theta-functions-have-to-do-with-quadratic-reciprocity What do theta functions have to do with quadratic reciprocity? Aleksandar Bahat 2013-01-28T00:47:24Z 2013-01-28T17:15:53Z <p>The <em>theta function</em> is the analytic function $\theta:U\to\mathbb{C}$ defined on the (open) right half-plane $U\subset\mathbb{C}$ by $\theta(\tau)=\sum_{n\in\mathbb{Z}}e^{-\pi n^2 \tau}$. It has the following important transformation property.</p> <blockquote> <p><strong>Theta reciprocity</strong>: $\theta(\tau)=\frac{1}{\sqrt{\tau}}\theta\left(\frac{1}{\tau}\right)$.</p> </blockquote> <p>This theorem, while fundamentally analytic&mdash;the proof is just Poisson summation coupled with the fact that a Gaussian is its own Fourier transform&mdash;has serious arithmetic significance.</p> <ul> <li><p>It is the key ingredient in the proof of the functional equation of the Riemann zeta function.</p></li> <li><p>It expresses the <em>modularity</em> of the theta function (of course, $\theta$ is not literally a modular form, since it is not even defined on all of the upper half-plane, but a simple change of variables can fix that).</p></li> </ul> <p>Theta reciprocity also provides an analytic proof (actually, the <em>only</em> proof, as far as I know) of the Landsberg-Schaar relation</p> <p>$$\frac{1}{\sqrt{p}}\sum_{n=0}^{p-1}\exp\left(\frac{2\pi i n^2 q}{p}\right)=\frac{e^{\pi i/4}}{\sqrt{2q}}\sum_{n=0}^{2q-1}\exp\left(-\frac{\pi i n^2 p}{2q}\right)$$</p> <p>where $p$ and $q$ are arbitrary positive integers. To prove it, apply theta reciprocity to $\tau=2iq/p+\epsilon$, $\epsilon>0$, and then let $\epsilon\to 0$.</p> <p>This reduces to the formula for the quadratic Gauss sum when $q=1$:</p> <p>$$\sum_{n=0}^{p-1} e^{2 \pi i n^2 / p} = \begin{cases} \sqrt{p} &amp; \textrm{if } \; p\equiv 1\mod 4 \\ i\sqrt{p} &amp; \textrm{if } \; p\equiv 3\mod 4 \end{cases}$$</p> <p>(where $p$ is an odd prime). From this, it's not hard to deduce Gauss's "golden theorem".</p> <blockquote> <p><strong>Quadratic reciprocity</strong>: $\left(\frac{p}{q}\right)\left(\frac{q}{p}\right)=(-1)^{(p-1)(q-1)/4}$ for odd primes $p$ and $q$.</p> </blockquote> <p>For reference, this is worked out in detail in the paper "<a href="http://www.math.kth.se/~akarl/langmemorial.pdf" rel="nofollow">Applications of heat kernels on abelian groups: $\zeta(2n)$, quadratic reciprocity, Bessel integrals</a>" by Anders Karlsson.</p> <hr> <p>I feel like there is some deep mathematics going on behind the scenes here, but I don't know what.</p> <blockquote> <p>Why should we expect theta reciprocity to be related to quadratic reciprocity? Is there a high-concept explanation of this phenomenon? If there is, can it be generalized to other reciprocity laws (like Artin reciprocity)?</p> </blockquote> <p>Hopefully some wise number theorist can shed some light on this!</p> http://mathoverflow.net/questions/117968/can-we-categorify-the-formula-for-the-quadratic-gauss-sum Can we categorify the formula for the quadratic Gauss sum? Aleksandar Bahat 2013-01-03T15:01:39Z 2013-01-03T15:01:39Z <h2>Background</h2> <p>Fix an odd prime $p$ and set $\zeta=e^{2\pi i/p}$. We define the <em>quadratic Gauss sum</em> as</p> <p>$$g=\sum_{n=0}^{p-1} \zeta^{n^2}.$$</p> <p>It's pretty easy to show that</p> <p>$$g^2= \begin{cases} p &amp; \textrm{if } p\equiv 1 \mod 4 \\ -p &amp; \textrm{if } p\equiv 3 \mod 4, \end{cases}$$</p> <p>and from this we can deduce quadratic reciprocity; it's harder to determine the modulus. We can actually find an explicit formula for $g$, namely:</p> <p>$$g= \begin{cases} \sqrt{p} &amp; \textrm{if } p\equiv 1 \mod 4 \\ i\sqrt{p} &amp; \textrm{if } p\equiv 3 \mod 4. \end{cases}$$</p> <p>This is the result I refer to for the remainder of the question.</p> <h2>Question</h2> <blockquote> <p><strong>Can we categorify this result?</strong></p> </blockquote> <p>By <em>categorification</em>, I mean the opposite of <em>decategorification</em>, and by decategorification, I mean the process of removing structure by e.g. taking the cardinality of a set or the dimension of a vector space. (Thus an example of categorification would be interpreting some combinatorial identity of positive integers as a bijection between sets.) This is intentionally vague, because there are plenty of people who have a much better idea of what constitutes categorification than I do, so feel free to interpret "categorification" liberally.</p> <h2>Motivation</h2> <p>Gauss's original proof of our result uses <a href="http://en.wikipedia.org/wiki/Gaussian_binomial_coefficient" rel="nofollow">q-binomial coefficients</a>. (A modern exposition of this proof can be found in <a href="http://www.ams.org/journals/bull/1981-05-02/S0273-0979-1981-14930-2/" rel="nofollow">"The determination of Gauss sums"</a> by Bruce C. Berndt and Ronald J. Evans.)</p> <p>Now, $q$-binomial coefficients can be categorified by Grassmannian varieties. What I mean by that is: the $q$-binomial coefficient $\binom{n}{k}_q$ is the number of $k$-dimensional subspaces of an $n$-dimensional vector space over the finite field $\mathbb{F}_q$, i.e. the cardinality of the Grassmannian $\textrm{Gr}(n,k)$. Basically, I'm wondering if there is some way this can be connected to the formula for the quadratic Gauss sum, seeing as how the formula is clearly related to the properties of $q$-binomial coefficients.</p> http://mathoverflow.net/questions/117668/new-grand-projects-in-contemporary-math/117819#117819 Answer by Aleksandar Bahat for New grand projects in contemporary math Aleksandar Bahat 2013-01-02T00:01:05Z 2013-01-02T00:01:05Z <p>The field with one element $\mathbb{F}_1$ (a.k.a. F-un).</p> <p>Having a precise notion of such a field would allow us to further exploit the analogy between number fields and function fields (much like discrete valuation rings, Dedekind domains, schemes, etc. have done in the past). In particular, with a suitable notion of $\mathbb{F}_1$ we should be able to find a proof of the Riemann hypothesis based on Weil’s proof of the Riemann hypothesis for curves over finite fields.</p> <p>The integers are not an algebra over any field in the classical sense, which makes it a priori impossible to adapt Weil's argument to this case. For this analogy to work, a good definition of $\mathbb{F}_1$ should have the property that $\mathbb{Z}$ is an $\mathbb{F}_1$-algebra.</p> <p>Furthermore, there are lots of connections with “q-deformations”. I don't know much about this, but John Baez has some nice stuff written at <a href="http://math.ucr.edu/home/baez/twfcontents.html" rel="nofollow"><em>This Week's Finds in Mathematical Physics</em></a>, weeks 183-187.</p> <p>An example of the cool unification we can do with $\mathbb{F}_1$ at our disposal: we can make sense of statements like</p> <blockquote> <p>Combinatorics is linear algebra over $\mathbb{F}_1$.</p> </blockquote> <p>(This idea is originally due to Jacques Tits, if I recall correctly.)</p> <p>For a possible way of rigorously developing such a theory, see Nikolai Durov's paper <a href="http://arxiv.org/abs/0704.2030" rel="nofollow">New Approach to Arakelov Geometry</a>.</p> http://mathoverflow.net/questions/105038/continuous-notions-with-compelling-discrete-analogues/105057#105057 Answer by Aleksandar Bahat for Continuous notions with compelling discrete analogues Aleksandar Bahat 2012-08-19T21:21:06Z 2012-08-19T22:02:02Z <p>One of my favorite examples of this is the "q-calculus", which is like a multiplicative version of the classical subject of calculus of finite differences. One can, using suitably defined "q" versions of the derivative, integral, and so on, recover analogues of most of the usual theorems in calculus. But what's more interesting is that this all ties in with noncommutative geometry and the field with one element (see John Baez's This Weeks Finds in Mathematical Physics).</p> http://mathoverflow.net/questions/120067/what-do-theta-functions-have-to-do-with-quadratic-reciprocity/120129#120129 Comment by Aleksandar Bahat Aleksandar Bahat 2013-02-07T16:57:56Z 2013-02-07T16:57:56Z Correction: that should read &quot;generating functions&quot; rather than &quot;generated functions&quot;. http://mathoverflow.net/questions/120906/algebraic-number-theory-building-and-simplifying Comment by Aleksandar Bahat Aleksandar Bahat 2013-02-06T06:06:09Z 2013-02-06T06:06:09Z The problem is, however, that schemes and cohomology could just as well fall under the &quot;building new things&quot; category... http://mathoverflow.net/questions/120906/algebraic-number-theory-building-and-simplifying Comment by Aleksandar Bahat Aleksandar Bahat 2013-02-06T06:05:05Z 2013-02-06T06:05:05Z I'm not sure if this counts as a &quot;simplification&quot; per se (which is why I'm posting it as a comment), but certainly arithmetic geometry can make some number theoretic results more &quot;natural&quot;, at least for someone like me who likes to think geometrically. Although schemes and cohomology may not fit your criterion of accessibility to undergraduates, they could rightfully be considered a means of simplifying parts of algebraic number theory. http://mathoverflow.net/questions/120485/fourier-analytic-proofs/120601#120601 Comment by Aleksandar Bahat Aleksandar Bahat 2013-02-02T21:09:47Z 2013-02-02T21:09:47Z I believe the first appearance of this proof is in the paper &quot;Uber die Gausschen Summen&quot; by Issai Schur. http://mathoverflow.net/questions/120067/what-do-theta-functions-have-to-do-with-quadratic-reciprocity/120129#120129 Comment by Aleksandar Bahat Aleksandar Bahat 2013-01-31T15:50:57Z 2013-01-31T15:50:57Z I suspected there was something Langlands-y about this. That's probably the best &quot;big picture&quot; explanation. Unfortunately, I don't know enough about Langlands to get very far with this. I like what you're getting at in the last paragraph; Jacobi's proof is &quot;natural&quot; (in the sense that generated functions are natural), and so I guess it's not a big leap to try to generalize that. http://mathoverflow.net/questions/120067/what-do-theta-functions-have-to-do-with-quadratic-reciprocity/120071#120071 Comment by Aleksandar Bahat Aleksandar Bahat 2013-01-31T14:18:10Z 2013-01-31T14:18:10Z I've heard of Hecke's generalization before, but I still feel my &quot;Why should we expect...&quot; question is unresolved. Although the usefulness of analytic functions in number theory is no longer surprising to me, I'd like to understand specifically why somebody would see quadratic reciprocity and think, &quot;Hmm, $\theta(z)=z^{-1/2}\theta(1/z)$ is relevant.&quot; Why this piece of analysis in particular? http://mathoverflow.net/questions/120067/what-do-theta-functions-have-to-do-with-quadratic-reciprocity Comment by Aleksandar Bahat Aleksandar Bahat 2013-01-28T21:12:54Z 2013-01-28T21:12:54Z Interesting! I didn't know about Dedekind sums. I'll have to read up on this some time soon. Yet another piece of the grand puzzle... http://mathoverflow.net/questions/117968/can-we-categorify-the-formula-for-the-quadratic-gauss-sum Comment by Aleksandar Bahat Aleksandar Bahat 2013-01-05T00:48:50Z 2013-01-05T00:48:50Z Thanks for your comment. I've heard of that result before somewhere, but never managed to track down a reference. It's very interesting, but doesn't really answer my question. If I had some result about objects that decategorified to this Gauss sum identity, then this paper might give me a proof &quot;for free&quot;; the problem is, I don't know <i>what</i> to prove about the objects yet! http://mathoverflow.net/questions/117668/new-grand-projects-in-contemporary-math/117819#117819 Comment by Aleksandar Bahat Aleksandar Bahat 2013-01-02T17:29:43Z 2013-01-02T17:29:43Z Also, thank you for pointing that out from the paper. There is definitely still a long way to go before we &quot;really find&quot; $\mathbb{F}_1$, but from what I know, this is one of the most substantial theoretical developments so far. http://mathoverflow.net/questions/117668/new-grand-projects-in-contemporary-math/117819#117819 Comment by Aleksandar Bahat Aleksandar Bahat 2013-01-02T16:19:34Z 2013-01-02T16:19:34Z Well, if Durov's definition solved the Riemann hypothesis, I think we would have heard! http://mathoverflow.net/questions/105038/continuous-notions-with-compelling-discrete-analogues/105057#105057 Comment by Aleksandar Bahat Aleksandar Bahat 2012-08-19T22:02:38Z 2012-08-19T22:02:38Z You're right. I added a reference.